Question

我是Objective-C的新手，我需要操作与c ++代码相同的代码。

typedef long long ll;
vector< pair<ll, int> > v;
for(int i = 0; i < N; i++){
   v.push_back( make_pair(token[i], N - i));
}
sort(v.begin(), v.end());

我需要在Objective-C中使用相同的代码而不会丢失面向对象。

这是我尝试过的Objective-C

#import <Foundation/Foundation.h>

typedef long long ll;


@interface Main:NSObject
-(void) main;
@end

@implementation Main

-(void) main{

    struct Pair{
        ll first;
        int second;
    };

    NSMutableArray *v = [NSMutableArray new];
    int N = 3;
    NSMutableArray *vector = [NSMutableArray arrayWithCapacity:(int)N];

    int i;
    for(i=0;i<N;i++){

        ll value = (ll)i;
        printf("Value is: %lld",(ll)i);
        [vector addObject:[NSNumber numberWithLongLong:(value)]];
    }

    for(int i = 0; i < N; i++){

        struct Pair *p = malloc(sizeof(struct Pair));;
        ll firstValue = [[vector objectAtIndex:i] longLongValue];

        NSLog(@"\nFirst Value is: %lld",firstValue);

        p->first = firstValue;
        p->second =  (int)N - i;

        NSLog(@"\n\nAfter append it to P:%lld and Int is: %d \n\n",p->first,p->second);

        NSValue *value = [NSValue valueWithBytes:p objCType:@encode(struct Pair)];

        [v addObject:value];
    }


    [v sortedArrayUsingComparator:^NSComparisonResult(id  _Nonnull obj1, id  _Nonnull obj2) {

        struct Pair *first = (__bridge struct Pair *)obj1;
        struct Pair *second = (__bridge struct Pair *)obj2;

        NSLog(@"%lld %d",first->first,first->second);

        return first->first < second->second;
    }];

    for(int i = 0;  i < N; i++){

        //I need to print first and second from V
    }
}
@end


int main() {
    @autoreleasepool {

        Main *main = [[Main alloc]init];
        [main main];
    }
    return 0;
}

问题：

我成功到达NSLog(@"\n\nAfter append it to P:%lld and Int is: %d \n\n",p->first,p->second);但我无法从v NSLog(@"%lld %d",first->first,first->second);取回相同的值struct。

那么，如何从v

重新插入

export const ex: any;

是否用简单的方法解决了Objective-C的C ++代码而不是这个？

以上Objective-C代码的输出：

值为：0值为：1值为：2

第一个值是：0

将其追加到P：0并且Int为：3第一个值为：1

将其追加到P：1并且Int为：2第一个值为：2

将其附加到P：2并且Int为：1

80501841873530129 0

80501841873530129 0

Answer 1

您正在使用带字节的值从指针构造NSValue ..
您正在对NSValue进行排序，但是通过网桥将其作为Pair *进行投放..
您可以通过malloc在堆上进行分配，如果您希望代码与C ++版本完全相同（每个Pair作为副本存储在向量中，则不需要;}指针）。

如果你想使用指针（到原始内存），那么：

#import <Foundation/Foundation.h>

typedef long long ll;

@interface Main:NSObject
-(void) main;
@end

@implementation Main

-(void) main{

    struct Pair {
        ll first;
        int second;
    };

    NSMutableArray *v = [NSMutableArray new];
    int N = 3;
    NSMutableArray *vector = [NSMutableArray arrayWithCapacity:(int)N];

    int i;
    for(i=0;i<N;i++){

        ll value = (ll)i;
        printf("Value is: %lld",(ll)i);
        [vector addObject:[NSNumber numberWithLongLong:(value)]];
    }

    for(int i = 0; i < N; i++){

        struct Pair *p = malloc(sizeof(struct Pair));
        ll firstValue = [[vector objectAtIndex:i] longLongValue];

        NSLog(@"\nFirst Value is: %lld",firstValue);

        p->first = firstValue;
        p->second =  (int)N - i;

        NSLog(@"\n\nAfter append it to P:%lld and Int is: %d \n\n", p->first, p->second);

        //Store pointer p as NSValue.
        NSValue *value = [NSValue valueWithPointer:p];
        [v addObject:value];
    }


    [v sortedArrayUsingComparator:^NSComparisonResult(NSValue *obj1, NSValue *obj2) {

        //call `getValue` to get the pointer back..
        struct Pair *first;
        [obj1 getValue:&first];

        struct Pair *second;
        [obj2 getValue:&second];

        NSLog(@"%lld %d", first->first, first->second);

        return first->first < second->second;
    }];

    for(int i = 0;  i < N; i++) {
        NSValue *value = [v objectAtIndex:i];

        //Call `getValue` to get the pointer back..
        struct Pair *pair;
        [value getValue:&pair];

        NSLog(@"%lld %d", pair->first, pair->second);

        free(pair); //clean up..
    }
}
@end


int main() {
    @autoreleasepool {

        Main *main = [[Main alloc] init];
        [main main];
    }
    return 0;
}

如果你不想使用malloc和免费的东西或处理指针（原始内存）..那么：

#import <Foundation/Foundation.h>

typedef long long ll;

@interface Main:NSObject
-(void) main;
@end

@implementation Main

-(void) main{

    struct Pair {
        ll first;
        int second;
    };

    NSMutableArray *v = [NSMutableArray new];
    int N = 3;
    NSMutableArray *vector = [NSMutableArray arrayWithCapacity:(int)N];

    int i;
    for(i=0;i<N;i++){

        ll value = (ll)i;
        printf("Value is: %lld",(ll)i);
        [vector addObject:[NSNumber numberWithLongLong:(value)]];
    }

    for(int i = 0; i < N; i++){

        struct Pair p;
        ll firstValue = [[vector objectAtIndex:i] longLongValue];

        NSLog(@"\nFirst Value is: %lld",firstValue);

        p.first = firstValue;
        p.second =  (int)N - i;

        NSLog(@"\n\nAfter append it to P:%lld and Int is: %d \n\n", p.first, p.second);

        //Encode p as a struct into an `NSValue`
        NSValue *value = [NSValue value:&p withObjCType:@encode(struct Pair)];
        [v addObject:value];
    }


    [v sortedArrayUsingComparator:^NSComparisonResult(NSValue *obj1, NSValue *obj2) {

        //Get each pair back
        struct Pair first;
        [obj1 getValue:&first];

        struct Pair second;
        [obj2 getValue:&second];

        NSLog(@"%lld %d",first.first, first.second);

        return first.first < second.second;
    }];

    for(int i = 0;  i < N; i++) {
        NSValue *value = [v objectAtIndex:i];

        //print each pair..
        struct Pair pair;
        [value getValue:&pair];

        NSLog(@"%lld %d", pair.first,pair.second);
    }
}
@end


int main() {
    @autoreleasepool {

        Main *main = [[Main alloc] init];
        [main main];
    }
    return 0;
}

Objective-C如何使用矢量对？

1 个答案: